TITLE: Find the limit of $4^n\cdot\binom{2n}{n}/\binom{4n}{2n}$
QUESTION [5 upvotes]: I am trying to prove that $$f(n)=4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}$$ converges as $n\rightarrow\infty$.
I have already tried to use the fact that, if $n, k \in\mathbb{N}, n\geq k\geq1,$ then $(\frac{n}{k})^k\leq\binom{n}{k}\leq e^k(\frac{n}{k})^k$.  However, this has been to no avail.
Could anybody suggest how I could approach this proof?

REPLY [3 votes]: Inequality $(9)$ from this answer shows that
$$
\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}
$$
Therefore,
$$
\sqrt{\frac{2n+\frac14}{n+\frac13}}\le4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}\le\sqrt{\frac{2n+\frac13}{n+\frac14}}
$$
The Squeeze Theorem says that
$$
\lim_{n\to\infty}4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}=\sqrt2
$$